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The 2-adic absolute value is the p-adic absolute value for the prime number $p=2$. It is defined on the rational numbers by $|x{|}_2=2^{-v_2(x)}$ for $x\ne 0$ and $|0{|}_2=0$, and extends to their P-adic number completion ${\mathbb{Q}}_2$. For a nonzero integer, count factors of two until the remaining factor is odd; for a rational number, subtract the denominator's count from the numerator's count.

The associated metric reverses a familiar comparison:Although $13$ is eight units from $5$ on the ordinary number line, it agrees with $5$ modulo $8$, whereas $6$ already disagrees modulo $2$. The 2-adic absolute value measures that agreement. The value $|3/8{|}_2=8$ also shows why this Non-Archimedean absolute value must be distinguished from the ordinary absolute value $|3/8|=3/8$.

In the left panel, differences farther from the diagonal grow in ordinary absolute value. In the right panel, nonzero even differences are closer than odd differences, and each further factor of two halves the 2-adic absolute value. The highlighted pairs have opposite distance rankings. The repeated pattern is the nested congruence class structure of an ultrametric space, visible without adding any new points.

Let $L/K$ be a finite Galois extension of a field complete for a Non-Archimedean absolute value. Suppose $\alpha$ is uniquely closest to $\beta$ among the roots of a polynomial over $K$ lying in $L$. Every element of $\mathrm{Gal}(L/K(\beta ))$ preserves absolute values on a field and permutes these roots. It sends $\alpha$ to an equally close root, hence fixes $\alpha$. The Fundamental theorem of Galois theory gives $K(\alpha )\subseteq K(\beta )$. The converse inclusion need not hold: for roots $0,\sqrt{p},-\sqrt{p}$, the element $p\sqrt{p}$ is uniquely closest to zero but does not belong to ${\mathbb{Q}}_p$.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension of an absolute value to every finite extension. Factor a monic $g\in {\mathcal{O}}_K[X]$ into monic irreducible factors over $K$. All factors have integral coefficients because their roots have absolute value at most one. By pure-power reduction of a monic irreducible polynomial, each factor reduces to a power of a single irreducible polynomial. If $\overline{g}={\overline{g}}_1{\overline{g}}_2$ is a coprime monic factorization, assign each whole irreducible factor of $g$, with its multiplicity, to the unique side containing its residue factor. The resulting monic $g_1,g_2\in {\mathcal{O}}_K[X]$ satisfy $g=g_1{g}_2$ and reduce to the prescribed factors. Completeness of $K$ is not required for this proof.

For a locally compact field with a nontrivial Non-Archimedean absolute value, scaling a compact neighborhood makes its valuation ring compact. Its open maximal ideal gives a finite discrete residue field, so that ideal is also closed and compact. A nonzero element of maximal absolute value in the ideal generates it. Successive division by this uniformizer makes the value group discrete.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Let $K$ be complete for a Non-Archimedean absolute value and let $V$ be a finite-dimensional vector space with a norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$ and the ultrametric inequality. All such norms are equivalent to the maximum of coordinates in any basis. The upper bound is immediate. For the lower bound, induct on dimension. The span $W$ of all but the last basis vector is complete by induction, hence closed for the given norm. Thus $\delta ={\inf }_{w\in W}\Vert e_n-w\Vert >0$. For $v=w+a{e}_n$, $\Vert v\Vert \ge |a|\delta$; also $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e_n\Vert )$. The induction bound on coordinates of $w$ now bounds all coordinates by a constant times $\Vert v\Vert$. This proof needs no local compactness.

A complete field with a nontrivial Non-Archimedean absolute value is locally compact exactly when its valuation is discrete and its residue field is finite. In that case its valuation ring is the inverse limit of the finite rings $\mathcal{O}/{\mathfrak{m}}^n$ and is compact. The nontrivial valuation in the local compactness criterion is essential.

For a field with a Non-Archimedean absolute value, define the coefficient norm $\Vert f{\Vert }_G={\max }_i|a_i|$ for $f(T)={\sum }_i{a}_i{T}^i$. Then $\Vert fg{\Vert }_G=\Vert f{\Vert }_G\Vert g{\Vert }_G$. For nonzero polynomials, scale each by one of its coefficients of maximum absolute value; the resulting polynomials have integral coefficients and at least one unit coefficient. Their reductions in the residue field are nonzero, so their product is nonzero. The product therefore also has coefficient norm $1$, proving the formula after reversing the scaling. This compares binary-polynomial coefficient heights with the heights of their roots.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value at most one. Some prime $p$ must have value less than one, or the value would be trivial on every rational number. Two distinct such primes are impossible by a Bezout identity. Prime factorization then gives the displayed formula. Distinct primes give inequivalent topologies, since powers of $p$ tend to zero only for the $p$-adic absolute value among these values.

Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

For a finite extension of fields complete for a Non-Archimedean absolute value, use the unique extended field absolute value. The extension is unramified when its residue extension is a separable field extension andEquivalently, it has ramification index one, a separable field extension of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.

Put $n=[L:K]=[k_L:k_K]$. Since $\bar{x}$ generates the residue extension, $1,\bar{x},\dots ,{\bar{x}}^{n-1}$ are linearly independent over $k_K$. Their lifts $1,x,\dots ,x^{n-1}$ are linearly independent over $K$: a nonzero relation could be divided by a coefficient of largest field absolute value, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-basis of $L$.

For any $y={\sum }_{j=0}^{n-1}{a}_j{x}^j$, choose $a_r$ of largest field absolute value. All coefficients of $y/a_r$ belong to ${\mathcal{O}}_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue basis, so $y/a_r$ is a unit of ${\mathcal{O}}_L$. This proves the residue-basis norm formula for an unramified extensionHence $y\in {\mathcal{O}}_L$ exactly when every $a_j\in {\mathcal{O}}_K$. We concludeThe last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a uniformizer nor discrete valuation, and applies to every residue generator specified in the question.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.

For a nonzero Laurent series let ${\mathrm{ord}}_t$ be its least exponent. Choose $|x|=p^{-{\mathrm{ord}}_t(x)}$ and $|0|=0$. This is a multiplicative Non-Archimedean absolute value, and its valuation ring is ${\mathbb{F}}_p[[t]]$.

TakeIf ${\gamma }_1,\dots ,{\gamma }_f$ is an ${\mathbb{F}}_p$-basis of ${\mathbb{F}}_{p^f}$, the elements ${\gamma }_i{u}^j$, $1\le i\le f$, $0\le j<e$, form a $K$-basis of $L$. To see this, group the exponents of each Laurent series modulo $e$ and then expand its coefficients in the constant-field basis. Uniqueness of those two expansions proves linear independence. Thus $[L:K]=ef$.

Its valuation ring is ${\mathbb{F}}_{p^f}[[u]]$, with maximal ideal $(u)$ and residue field ${\mathbb{F}}_{p^f}$. Moreover ${\mathrm{ord}}_u(t)=e$, so the ramification index is $e$ and the residue degree is $f$. The unique extending absolute value is explicitlyThis restricts to the chosen absolute value because ${\mathrm{ord}}_u(x)=e{\mathrm{ord}}_t(x)$ on $K$. The prescribed ramification and residue degree for a Laurent series field construction works even when $p\mid e$; separability is not required in the question.

A Non-Archimedean absolute value on a field is a function $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ with $|x|=0$ exactly for $x=0$, $|xy|=|x||y|$, and the ultrametric inequality $|x+y|\le \max (|x|,|y|)$. It is nontrivial if some nonzero element has value different from one. Equivalent absolute values induce the same topology; for nontrivial absolute values this is equivalent to $|x{|}_2=|x{|}_1^c$ for some $c>0$.

Here is also a proof of the latter equivalence in the case needed. A positive power plainly preserves the open neighborhoods of zero. Conversely, if the topologies coincide, then $|x{|}_1<1$ if and only if $x^n\to 0$ in the first topology, hence if and only if $|x{|}_2<1$. Choose $a$ with $0<|a{|}_1<1$. For every $x\ne 0$, $m\in \mathbb{Z}$ and $n\ge 1$, whether $|x^n/a^m|<1$ agrees for the two values. Comparison with all rational numbers $m/n$ therefore identifies the two ratios $\log |x{|}_j/\log |a{|}_j$. Thus $|x{|}_2=|x{|}_1^c$ with $c=\log |a{|}_2/\log |a{|}_1>0$.

Now let $|\cdot |$ be nontrivial and non-Archimedean on $\mathbb{Q}$. Since $|1|=1$ and every positive integer is a sum of ones, $|n|\le 1$ for every integer. If all nonzero integers had value one, multiplicativity would make the absolute value trivial on all rationals. Hence some integer has value less than one; its prime factorization shows that some prime $p$ has $|p|<1$.

There cannot be two distinct such primes $p,q$. A Bezout identity $ap+bq=1$ with $a,b\in \mathbb{Z}$ would giveThus every other prime has value one. Prime factorization of the numerator and denominator of $x\in {\mathbb{Q}}^{\times }$ now givesConversely $|x{|}_p=p^{-v_p(x)}$ is multiplicative and satisfies the ultrametric inequality, since $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Distinct primes give inequivalent values: $p^m\to 0$ for the $p$-adic absolute value but $|p^m{|}_q=1$ for $q\ne p$. We have proved the non-Archimedean part of Ostrowski theorem:There is exactly one equivalence class for each prime.

Two absolute values on a field are equivalent absolute values when they induce the same topology. For nontrivial absolute values on a field this is equivalent tofor a fixed real number $c>0$. The trivial absolute value is equivalent only to itself.

Restriction of a Non-Archimedean absolute value plainly preserves the ultrametric inequality. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either field, has absolute value on a field at most one. In particular every binomial coefficient has absolute value on a field at most one. The triangle inequality applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,Raise both sides to the power $1/N$ and let $N\to \infty$. Since $(N+1{)}^{1/N}\to 1$, this yieldsThis is the bounded-integer criterion for a non-Archimedean absolute value. The proof works for arbitrary field extensions, not just algebraic ones, and in positive characteristic as well.

First include the trivial absolute value if that convention is allowed. For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value on a field at most one, and some nonzero integer has absolute value on a field less than one: otherwise every nonzero rational number, being a quotient of two integers, would have absolute value on a field one. Factoring that integer shows that some prime number $p$ has $|p|<1$.

There cannot be two such prime numbers. If $|p|<1$ and $|q|<1$ for distinct prime numbers, choose integers $a,b$ with $ap+bq=1$. Then the ultrametric inequality givesEvery other prime number therefore has absolute value on a field one. Factoring numerator and denominator of a rational number givesThus the nontrivial equivalence classes are precisely the p-adic absolute values, one for each prime number.

Now let $F$ be a number field, with ring of integers of a number field $A$. If the restriction of an absolute value on a field to $\mathbb{Q}$ is trivial, a monic polynomial over $\mathbb{Q}$ satisfied by $x\in F$ forces $|x|\le 1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne 0$. Thus a nontrivial absolute value on a field defined on $F$ has a nontrivial restriction, associated with some prime number $p$.

Every $a\in A$ has $|a|\le 1$, by applying the same leading-term argument to its monic integral equation. Consequentlyis a proper prime ideal: it is an ideal by the ultrametric inequality, and $|ab|<1$ with $|a|,|b|\le 1$ implies that at least one factor has absolute value on a field less than one. It is nonzero because it contains $p$.

The localization $A_{\mathfrak{p}}$ lies in the valuation ring of the absolute value on a field: a denominator outside $\mathfrak{p}$ has absolute value on a field one. Since $A$ is a Dedekind domain, $A_{\mathfrak{p}}$ is a discrete valuation ring. Choose its uniformizer $\pi$. Each $x\in F^{\times }$ has a unique expressionBoth $u$ and $u^{-1}$ lie in the valuation ring, so $|u|=1$. Also $\pi$ belongs to its maximal ideal, hence $|\pi |<1$. Therefore $|x|=|\pi {|}^{v_{\mathfrak{p}}(x)}$.

Conversely each nonzero prime ideal of $A$ gives the Non-Archimedean absolute value $|x|=c^{v_{\mathfrak{p}}(x)}$ for any $0<c<1$, using the discrete valuation of $A_{\mathfrak{p}}$. All choices of $c$ are equivalent, and different prime ideals are distinguished by the elements of $A$ having absolute value on a field less than one. We have proved Non-Archimedean absolute values on a number field:The trivial class is additional when trivial absolute values on a field are admitted.

The intended statement uses a nontrivial Non-Archimedean absolute value. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N{\mathcal{O}}_K$ is contained in that ball. The valuation ring is closed, so $t^N{\mathcal{O}}_K$ is a closed subset of $C$ and is compact. Scaling shows that ${\mathcal{O}}_K$ itself is compact.

Its maximal ideal $\mathfrak{m}=\{x:|x|<1\}$ is open. The residue field $k={\mathcal{O}}_K/\mathfrak{m}$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak{m}$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak{m}$ is also closed and compact.

The continuous absolute value on a field attains its maximum on $\mathfrak{m}$. Nontriviality supplies a nonzero element there, so this maximum is $\rho =|\pi |$ for some $\pi \ne 0$, with $0<\rho <1$. If $x\in \mathfrak{m}$, then $|x/\pi |\le 1$, so $\mathfrak{m}=\pi {\mathcal{O}}_K$.

For arbitrary $x\ne 0$, choose the integer $n$ for which ${\rho }^{n+1}<|x|\le {\rho }^n$. Then $\rho <|x/{\pi }^n|\le 1$. This element cannot lie in $\mathfrak{m}$, by maximality of $\rho$, so its absolute value on a field is one. Therefore $|x|={\rho }^n$. This proves discrete valuation from nontrivial local compactness:

The nontrivial valuation in the local compactness criterion is essential. If the trivial absolute value is admitted, the displayed source claim is false: take $\mathbb{Q}$ with that absolute value on a field. It is complete and discrete, hence locally compact, but its valuation ring and residue field are both $\mathbb{Q}$, which is infinite. The intended nontrivial case has been proved above.

Every Non-Archimedean absolute value on $\mathbb{Q}$ is either the trivial absolute value, or $|\cdot {|}_p^c$ for a unique prime $p$ and some $c>0$. Here $|p{|}_p=p^{-1}$.

To prove this classification, the ultrametric inequality gives $|n|\le 1$ for all integers $n$. If every nonzero integer has absolute value one, multiplicativity makes the absolute value on a field trivial on all rational numbers. Otherwise some positive integer has absolute value less than one, and its prime factorization gives a prime $p$ with $0<|p|<1$. There cannot be two such primes: if $p\ne q$ and $ap+bq=1$ with integers $a,b$, then $1\le \max (|p|,|q|)<1$, a contradiction. Every other prime consequently has absolute value one. Unique prime factorization now givesConversely these are Non-Archimedean absolute values, by the valuation inequality $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Thus $\boxed{|\cdot |=|\cdot {|}_p^c(c>0)\text{ or the trivial value}}$. This is the non-Archimedean part of Ostrowski theorem, including the trivial case.